Respuesta :
Answer:
99% confidence interval for the proportion of lids produced that day with this defect somewhere between ( .0213 , .0987) or between [tex]2.13\%[/tex] to [tex]9.87\%[/tex]
Step-by-step explanation:
Given -
Sample size ( n) = 250
Sample proportion [tex](\widehat{p})[/tex] = [tex]\frac{15}{250}[/tex] = .06
[tex]\alpha = 1 -[/tex] confidence interval = 1 - .99 = .01
[tex]z_{\frac{\alpha}{2}}[/tex] = [tex]z_{\frac{.01}{2}}[/tex] = 2.58
99% confidence interval for the proportion of lids produced that day with this defect = [tex]\widehat{p} \pm z_{\frac{\alpha}{2}} \sqrt{\frac{(\widehat{p}) (1 - \widehat{p}) }{n}}[/tex]
= [tex].06 \pm z_{\frac{.01}{2}} \sqrt{\frac{(.06) (1 - .06) }{250}}[/tex]
= [tex].06 \pm 2.58\times .0150[/tex]
= [tex].06 \pm .0387[/tex]
= ( .06 + .0387 ) , ( .06 - .0387 )
= ( .0987 , .0213 )
